The second linear function is y = -6x + 8. Which value represents the possible difference between the two functions if x = 5 ?

Mathematics

2 answers:

ss7ja [257]3 years ago

8 0

ITS C AND IM DOING THIS TYPE OF WORK TOO. ITS MADDD HARD

ExtremeBDS [4]3 years ago

6 0

The answer is C)-22

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The largest possible circle is cut out of a square whose side length is 8 feet. What will be the approximate area, in square fee

valentina_108 [34]

<h3>Therefore the area of remaining board =13.76 square feet</h3>

Step-by-step explanation:

Given , The length of side of the square is 8 feet.

Since a circle is inscribed in the square. Then the diameter of the circle is equal to the length of side of the square .

Therefore the diameter of the circle is = 8 feet.

Radius of the circle is(r) = $\frac{8}{2}$ feet = 4 feet

The area of the circle is= 3.14 r²

= 3.14 × 4² square feet

= 50.24 square feet

The area of the square is = side × side

= 8×6 square feet

=64 square feet

Therefore the area of remaining board = (64- 50.24)square feet

=13.76 square feet

4 0

3 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$